Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.
Chapter 7.2, Problem 11P
(a)
To determine
The z intervals from x interval 4.5<x.
(a)
Expert Solution
Answer to Problem 11P
Solution: The z intervals from x interval is (−1.0<z).
Explanation of Solution
μ=4.8,σ=0.3x=4.5
We use the formula for normal distribution:
z=x−μσz=4.5−4.80.3z=−1.0
P(4.5<x)=P(−1.0<z)
(b)
To determine
The z intervals from x interval x<4.2.
(b)
Expert Solution
Answer to Problem 11P
Solution: The z intervals from x interval is z<−2.0.
Explanation of Solution
μ=4.8,σ=0.3x=4.2
We use the formula for normal distribution:
z=x−μσz=4.2−4.80.3z=−2.0
P(x<4.2)=P(z<−2.0)
(c)
To determine
The z intervals from x interval 4.0<x<5.5.
(c)
Expert Solution
Answer to Problem 11P
Solution: The z intervals from x interval is (−2.67<z<2.33).
Explanation of Solution
μ=4.8,σ=0.3x1=4.0,x2=5.5
We use the formula for normal distribution:
z=x−μσz1=4.0−4.80.3=−2.67z2=5.5−4.80.3=2.33
P(4.0<x<5.5)=P(−2.67<z<2.33)
(d)
To determine
The x intervals from z interval z<−1.44.
(d)
Expert Solution
Answer to Problem 11P
Solution: The x intervals from z interval is x<4.37.
Explanation of Solution
μ=4.8,σ=0.3z=−1.44
We use the formula for normal distribution:
z=x−μσx=−1.44(0.3)+4.8x=4.37
P(z<−1.44)=P(x<4.37)
(e)
To determine
The x intervals from z interval 1.28<z.
(e)
Expert Solution
Answer to Problem 11P
Solution: The x intervals from z interval is 5.18<x.
Explanation of Solution
μ=4.8,σ=0.3z=1.28
We use the formula for normal distribution:
z=x−μσx=1.28(0.3)+4.8x=5.18
P(1.28<z)=P(5.18<x)
(f)
To determine
The x intervals from z interval −2.25<z<−1.00.
(f)
Expert Solution
Answer to Problem 11P
Solution: The x intervals from z interval is 4.13<x<4.50.
Whether a female having a RBC count of 5.9 or higher would be considered unusually high.
(g)
Expert Solution
Answer to Problem 11P
Solution: Yes, female having a RBC count of 5.9 or higher would be considered unusually high.
Explanation of Solution
μ=4.8,σ=0.30x=5.9
We use the formula for normal distribution:
z=x−μσz=5.9−4.80.30z=3.67
According to Figure 6-15, 99.7% of the data values lie within 3 standard deviation of the means. Since the obtained z-value is 3.67 is above three standard deviation of means, hence we can say that female having a RBC count of 5.9 or higher would be considered unusually high.
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